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Let $\newcommand\Com{\mathsf{Com}^H}\newcommand\Vect{\mathsf{Vect}}\Com$ be the category of right $H$-comodules. This has sufficiently many injectives, so we can compute the right derived functors of the left exact functor $F=\hom_{\Com}(k,\mathord-):\Com\to\Vect$, where $K$ denotes the trivial, $1$-dimensiona comodule. We can write $\newcommand\Ext{\mathrm{Ext}}\Ext_{\Com}^p(k,V)=R^pF(V)$.

Notice that $F(V)=V^H$ for all $V\in\Com$.

If $$\tag{$\star$}0\to W\to V\to U\to0$$ is a short exact sequence in $\Com$, then we have a long exact sequence for the derived functors $R^pF$, which starts with $$0\to W^H\to V^H\to U^H\to\Ext^1_{\Com}(k,W)\to\cdots$$

We can conclude, then, as usual, that the map $V^H\to U^H$ is surjective if, for example $\Ext^1_{\Com}(k,W)=0$. This can happen for various reasons: one obvious one is that $W$ be an injective comodule. A draconiant version of this is the condition that $H$ be cosemisimple.

To say something more intelligent, one would probably need to know more details about your concrete situation, though.

Dually, we can use the functor $\newcommand\box{\mathbin{\Box^H}}G=(\mathord-)\box k$, the cotensor product with the trivial module $k$. Here the traditional notation for its derived functors is $\newcommand\Cotor{\mathrm{Cotor}^H}\Cotor_p(\mathord-,k)=R^pG(\mathord-)$. The long exact sequence for the derived functors of $G$ applied to the short exact sequence $(\star)$ is now
$$0\to W^H\to V^H\to U^H\to\Cotor_1(W,k)\to\cdots$$
and we see that for $V^H\to U^H$ to be surjective is enough that $W$ be coflat.